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<meta name="description" content="平时在练习扒谱的时候有注意到：当你在指板上找出相邻的两个音时，就可以把调的范围从12个缩小到2个。比如：当你找出2弦空弦与2弦1品是这个调的音时，这两个音要么是这个调的Mi-Fa，要么是这个调的Si-Do，当是Mi-Fa时这个调就是G调，当时Si-Do时就是C调。可以猜想：不必找出这个调的所有音就可以确定这个调。那么最少找出几个音可以确定一个调呢？要找出哪几个音呢？  注：本文的讨论仅限于自然调式">
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<meta name="twitter:description" content="平时在练习扒谱的时候有注意到：当你在指板上找出相邻的两个音时，就可以把调的范围从12个缩小到2个。比如：当你找出2弦空弦与2弦1品是这个调的音时，这两个音要么是这个调的Mi-Fa，要么是这个调的Si-Do，当是Mi-Fa时这个调就是G调，当时Si-Do时就是C调。可以猜想：不必找出这个调的所有音就可以确定这个调。那么最少找出几个音可以确定一个调呢？要找出哪几个音呢？  注：本文的讨论仅限于自然调式">



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          <h1 class="post-title" itemprop="name headline">最少通过几个音可以确定一个调？</h1>
        

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        <p>平时在练习扒谱的时候有注意到：当你在指板上找出相邻的两个音时，就可以把调的范围从12个缩小到2个。比如：当你找出2弦空弦与2弦1品是这个调的音时，这两个音要么是这个调的Mi-Fa，要么是这个调的Si-Do，当是Mi-Fa时这个调就是G调，当时Si-Do时就是C调。<br>可以猜想：<strong>不必找出这个调的所有音就可以确定这个调</strong>。那么最少找出几个音可以确定一个调呢？要找出哪几个音呢？</p>
<blockquote>
<p>注：本文的讨论仅限于自然调式</p>
</blockquote>
<a id="more"></a>
<h1 id="问题分析"><a href="#问题分析" class="headerlink" title="问题分析"></a>问题分析</h1><p>自然大调中的12的调的组成音如下:</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><span class="line">C  调: C   D   E   F   G   A   B</span><br><span class="line">bD 调: bD  bE  F   bG  bA  bB  C</span><br><span class="line">D  调: D   E   bG  G   A   B   bD</span><br><span class="line">bE 调: bE  F   G   bA  bB  C   D</span><br><span class="line">E  调: E   bG  bA  A   B   bD  bE</span><br><span class="line">F  调: F   G   A   bB  C   D   E</span><br><span class="line">bG 调: bG  bA  bB  B   bD  bE  F</span><br><span class="line">G  调: G   A   B   C   D   E   bG</span><br><span class="line">bA 调: bA  bB  C   bD  bE  F   G</span><br><span class="line">A  调: A   B   bD  D   E   bG  bA</span><br><span class="line">bB 调: bB  C   D   bE  F   G   A</span><br><span class="line">B  调: B   bD  bE  E   bG  bA  bB</span><br></pre></td></tr></table></figure>
<p>我们先把问题具体化，以C调为例。就是我们要从C调中找出几个音，然后这几个音在剩余11个调中都不同时存在，此问题可以抽象成如下的数学问题：</p>
<blockquote>
<p>有N个非空集合。对于每一个集合，其中的每一个元素，在其他N-1个集合中至少存在一个集合包含这个元素。求每个集合的(不是其他N-1个集合中的任何一个集合的子集的)最小长度的子集  </p>
</blockquote>
<h1 id="解答啊解答"><a href="#解答啊解答" class="headerlink" title="解答啊解答"></a>解答啊解答</h1><blockquote>
<p>下面给出解答过程，😝没有兴趣的可以直接转到文末查看结论。</p>
</blockquote>
<p>可以用<strong>无敌暴力穷举法</strong>对C调音名集合的全部子集进行验证(*￣︶￣)<br>代码如下：</p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment"># -*- coding: utf-8 -*-</span></span><br><span class="line"><span class="keyword">import</span> pprint</span><br><span class="line"><span class="keyword">from</span> itertools <span class="keyword">import</span> combinations</span><br><span class="line"></span><br><span class="line">DIATONIC_SCALE = (<span class="string">'C'</span>, <span class="string">'bD'</span>, <span class="string">'D'</span>, <span class="string">'bE'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'bG'</span>, <span class="string">'G'</span>, <span class="string">'bA'</span>, <span class="string">'A'</span>, <span class="string">'bB'</span>,<span class="string">'B'</span>)</span><br><span class="line">INTERVAL = (<span class="number">2</span>, <span class="number">2</span>, <span class="number">1</span>, <span class="number">2</span>, <span class="number">2</span>, <span class="number">2</span>)</span><br><span class="line"></span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">def</span> <span class="title">special</span><span class="params">()</span>:</span></span><br><span class="line">    <span class="string">"""special"""</span></span><br><span class="line">    major_scales = []</span><br><span class="line">    <span class="keyword">for</span> ind, main <span class="keyword">in</span> enumerate(DIATONIC_SCALE):</span><br><span class="line">        scale = DIATONIC_SCALE[ind:] + DIATONIC_SCALE[:ind]</span><br><span class="line">        major_scale = []</span><br><span class="line">        major_scales.append(major_scale)</span><br><span class="line">        i = <span class="number">0</span></span><br><span class="line">        major_scale.append(scale[<span class="number">0</span>])</span><br><span class="line">        <span class="keyword">for</span> itv <span class="keyword">in</span> INTERVAL:</span><br><span class="line">            i += itv</span><br><span class="line">            major_scale.append(scale[i])</span><br><span class="line">    <span class="keyword">for</span> scale <span class="keyword">in</span> major_scales:</span><br><span class="line">        s = scale[<span class="number">0</span>]</span><br><span class="line">        main = [s + <span class="string">' '</span>, s][len(s) == <span class="number">2</span>]</span><br><span class="line">        <span class="comment"># print(main + ' 调: ' + '  '.join([s + ' ', s][len(s) == 2] for s in scale))</span></span><br><span class="line">    out = []</span><br><span class="line">    <span class="keyword">for</span> i <span class="keyword">in</span> range(<span class="number">7</span>): </span><br><span class="line">        <span class="keyword">for</span> subset <span class="keyword">in</span> combinations(major_scales[<span class="number">0</span>], i):</span><br><span class="line">            <span class="keyword">for</span> superset <span class="keyword">in</span> major_scales[<span class="number">1</span>:]:</span><br><span class="line">                subset = set(subset)</span><br><span class="line">                <span class="keyword">if</span> subset.issubset(set(superset)):</span><br><span class="line">                    <span class="keyword">break</span></span><br><span class="line">            <span class="keyword">else</span>:</span><br><span class="line">                out.append(subset)</span><br><span class="line">    out = [sorted(s,key = <span class="keyword">lambda</span> i:DIATONIC_SCALE.index(i)) <span class="keyword">for</span> s <span class="keyword">in</span> out]</span><br><span class="line">    <span class="keyword">return</span> out </span><br><span class="line"></span><br><span class="line"></span><br><span class="line"><span class="keyword">if</span> __name__ == <span class="string">'__main__'</span>:</span><br><span class="line">    out = special()</span><br><span class="line">    pprint.pprint(out)</span><br></pre></td></tr></table></figure>
<p>输出的结果为：<br><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line">[[<span class="string">'C'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'D'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'C'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>],</span><br><span class="line"> [<span class="string">'D'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'G'</span>, <span class="string">'A'</span>, <span class="string">'B'</span>]]</span><br></pre></td></tr></table></figure></p>
<h1 id="结果分析"><a href="#结果分析" class="headerlink" title="结果分析"></a>结果分析</h1><p>输出结果中的每一行表示一种可能，也就是说我们<strong>最少需要找出3个音就可以确定一个调</strong>了。</p>
<p>仔细观察结果还可以发现，每一种可能中都含有<code>F</code>和<code>B</code>，找出<code>F</code>和<code>B</code>后至少再从剩下的5个音中随便找出1个就可了。也就是说假如你没有同时找出<code>F</code>和<code>B</code>，就算你找到了6个音也不能确定是什么调。这里有两个极端的例子。</p>
<ul>
<li>当你找出<code>F</code>和<code>B</code>，其他11个调中同时含有这两个音的还有bG调，而且C调的另外5个音与bG调的另外5个音都不相同，所以再找一个音就可以确定是什么调了。</li>
<li>比如C调与G调就只有<code>F</code>和<code>bG</code>不一样，就算你找出了除<code>F</code>之外的6个音也不能确定是C调。</li>
</ul>
<p>然后还有个规律就是<code>F</code>到<code>B</code> ，<code>B</code>到<code>F</code>都是相隔6个半音。<br><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line"><span class="string">'C'</span>, <span class="string">'bD'</span>, <span class="string">'D'</span>, <span class="string">'bE'</span>, <span class="string">'E'</span>, <span class="string">'F'</span>, <span class="string">'bG'</span>, <span class="string">'G'</span>, <span class="string">'bA'</span>, <span class="string">'A'</span>, <span class="string">'bB'</span>,<span class="string">'B'</span></span><br></pre></td></tr></table></figure></p>
<p><strong>结论：找出两个相差6个半音的音（即4级音和7级音），再至少找出一个音就可以确定调了</strong></p>
<p>更适合实践的方法是：在指板上找出两个相邻的音，看比这两个音中的较高把位的音高6个半音的音是不是这个调的音，如果是的话那么较高把位的这个音就是<code>Fa</code>，不是的话这个音就是<code>Do</code>（此时比这个音高5个半音的音会是这个调的音）。如此就可以确定调了</p>
<p>emmmmmmmm说了这么多，那么这个结论对实际找调有什么帮助呢。说实话一点luan用都没有，纯属没事找乐子：）</p>

      
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